Since any bipartite quantum system has a corresponding (reduced)
density matrix to describe it, we can study the bipartite quantum
entanglement degree through its density matrix. It's well known
that the density matrix is composed of the state vectors of linear
Hilbert space, so the density matrix has the property of linear
decomposition and linear superposition which is the unique
property of quantum states, therefore a density matrix cannot be
used to describe a classical state, it only can be used to
describe quantum state. Speaking concretely, there isn't quantum
correlation between classical states, so a superposition of
several classical states cannot create new state through linear
decomposition and recombination. On the contrary, if a system is
described by the density matrix \(\rho =\sum _ip_i\vert \psi _i\rangle \langle \psi _i\vert \) , the system's state isn't
limited to the range \(\lbrace  \vert \psi _i\rangle \rbrace \) . The measuring
result of the system can be any linear combination state
\(\psi ^{^{\prime }}\)  of the states \(\lbrace  \vert \psi _i\rangle \rbrace \)  which makes
\(tr(\rho \vert \psi ^{^{\prime }}\rangle \langle \psi ^{^{\prime }}\vert )\)  nonzero. This
is because the states \(\lbrace  \vert \psi _i\rangle \rbrace \)  in the density
matrix \(\rho \)  can be freely linearly decomposed and recombined to
create new states. This means the states in a density matrix are
quantum entangled together. So the entanglement present in a
density matrix is just the quantum entanglement. This means that
only the density matrix which can be expressed by a direct product
of the subsystem's density matrices \(\rho _{AB}=\rho _A\otimes \rho _B\)  doesn't contain quantum entanglement. Any other density
matrix, including the following form [1]}:
\(\rho _{AB}\,=\,\sum _{i}p_i\,\rho ^A_i\otimes \rho ^B_i\,,\) 
